Answer:
the lowest point of energy the the graph reaches
Answer:
11.9 g of nitrogen monoxide
Explanation:
We'll begin by calculating the number of mole in 6.75 g of NH₃. This can be obtained as follow:
Mass of NH₃ = 6.75 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 6.75 / 17
Mole of NH₃ = 0.397 mole
Next, we shall determine the number of mole of NO produced by the reaction of 0.397 mole of NH₃. This can be obtained as follow:
4NH₃ + 5O₂ —> 4NO + 6H₂O
From the balanced equation above,
4 moles of NH₃ reacted to produce 4 moles of NO.
Therefore, 0.397 mole of NH₃ will also react to produce 0.397 mole of NO.
Finally, we shall determine the mass of 0.397 mole of NO. This can be obtained as follow:
Mole of NO = 0.397 mole
Molar mass of NO = 14 + 16 = 30 g/mol
Mass of NO =?
Mass = mole × molar mass
Mass of NO = 0.397 × 30
Mass of NO = 11.9 g
Thus, the mass of NO produced is 11.9 g
a) First, to get ΔG°rxn we have to use this formula when:
ΔG° = - RT ㏑ K
when ΔG° is Gibbs free energy
and R is the constant = 8.314 J/mol K
and T is the temperature in Kelvin = 25 °C+ 273 = 298 K
and when K = 4.4 x 10^-2
so, by substitution:
ΔG°= - 8.314 * 298 *㏑(4.4 x 10^-2)
= -7739 J = -7.7 KJ
b) then, to get E° cell for a redox reaction we have to use this formula:
ΔE° Cell = (RT / nF) ㏑K
when R is a constant = 8.314 J/molK
and T is the temperature in Kelvin = 25°C + 273 = 298 K
and n = no.of moles of e- from the balanced redox reaction= 3
and F is Faraday constant = 96485 C/mol
and K = 4.4 x 10^-2
so, by substitution:
∴ ΔE° cell = (8.314 * 298 / 3* 96485) *㏑(4.4 x 10^-2)
= - 2.7 x 10^-2 V
Blood is a liquid substance flowing through the arteries and veins of the body. In a healthy normal person, approximately 45% of their blood volume are cells (mostly red blood cells, white blood cells and platelets).
hope this helps!
Answer: 1.56 ATM
Explanation: if we assume temperature is constant, gas obeys
Boyles law pV= constant. Then p1·V1= p2·V2. And V1 = p2V2/p1
= 3.0 atm·0,52 l / 1.0 atm
